Proof of The Tracial Algebra of a Noncommutative Law: a Norm-Closed Unital *-Algebra with a Faithful Positive Trace, Determined by Vacuum Vectors, Closed under Square Roots

Conventions. By The Complex Hilbert Completion is a Complex Hilbert Space Containing a Dense Isometric Image, and Bounded Complex-Linear Maps Extend to It §hilbert, $\mathcal{H}$ is a complex Hilbert space, so every $S\in\mathcal{L}(\mathcal{H})$ has an adjoint $S^{*}\in\mathcal{L}(\mathcal{H})$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint. By the definition of an adjoint in Adjoint of a Linear Map between Complex Inner Product Spaces §adjoint, and because $S$ is the adjoint of $S^{*}$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus,

Since $\lVert S\rVert_{\mathrm{op}}$ is a bound for $S$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound, linearity gives

All operators $L_{p}$, $R_{p}$ below lie in $\mathcal{L}(\mathcal{H})$ by Left and Right Multiplication Operators on the Complex GNS Space of a Noncommutative Law: Boundedness, Algebra Rules, Adjoints, Commutation, the Vacuum and the Conjugation §multiplication, and sums, scalar multiples and composites of elements of $\mathcal{L}(\mathcal{H})$, the zero map and $I$ lie in $\mathcal{L}(\mathcal{H})$, a complex vector space in which composition distributes over sums and commutes with scalar multiples, by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §operations.

Preliminary: approximation by classes. Let $\beta=\operatorname{Re}h_{\lambda}$. By The Complex GNS Space of a Tracial State on Noncommutative Polynomials §gns and The Hilbert Completion of a Complex Vector Space with a Positive Semidefinite Hermitian Form §completion, $\mathcal{H}$ is the Hilbert completion $H_{\beta}$ of $(\mathcal{P}_{d},\beta)$ with the same addition and multiplication by real scalars, and its norm is the norm $|\cdot|$ of $H_{\beta}$ by The Complex Hilbert Completion is a Complex Hilbert Space Containing a Dense Isometric Image, and Bounded Complex-Linear Maps Extend to It §hilbert; so $\mathcal{H}$ and $H_{\beta}$ carry the same metric. Let $\xi\in\mathcal{H}$. By The Hilbert Completion of a Real Vector Space with a Positive Semidefinite Symmetric Bilinear Form §completion, $\xi=[u]$ for a $\beta$-Cauchy sequence $u=(u_{k})_{k\in\mathbb{N}}$ in $\mathcal{P}_{d}$, and by The Hilbert Completion is a Real Hilbert Space Containing a Dense Isometric Image, and Bounded Linear Maps Extend to It §dense the sequence $(J_{\beta}u_{k})$ converges to $\xi$. Since $J_{\beta}u_{k}=J_{h_{\lambda}}u_{k}=\widehat{u_{k}}$ by The Hilbert Completion of a Complex Vector Space with a Positive Semidefinite Hermitian Form §canonical-map and The Complex GNS Space of a Tracial State on Noncommutative Polynomials §classes, the classes $\widehat{u_{k}}$ converge to $\xi$ in $\mathcal{H}$, that is, $\lVert\widehat{u_{k}}-\xi\rVert\to0$.

(P1) If $S,S'\in\mathcal{L}(\mathcal{H})$ and $S\widehat{q}=S'\widehat{q}$ for every $q\in\mathcal{P}_{d}$, then $S=S'$. Let $\xi\in\mathcal{H}$ and let $(\widehat{u_{k}})$ converge to $\xi$ as above. By $(\ast)$, $\lVert S\widehat{u_{k}}-S\xi\rVert\le\lVert S\rVert_{\mathrm{op}}\lVert\widehat{u_{k}}-\xi\rVert\to0$, so $(S\widehat{u_{k}})$ converges to $S\xi$; likewise $(S'\widehat{u_{k}})$ converges to $S'\xi$. These are the same sequence, so $S\xi=S'\xi$ by Uniqueness of Limits in a Metric Space.

(P2) If $\zeta\in\mathcal{H}$ and $\langle\zeta,\widehat{q}\rangle=0$ for every $q\in\mathcal{P}_{d}$, then $\zeta=0$. Let $(\widehat{u_{k}})$ converge to $\zeta$ as above. By conditions 2 and 3 of Complex Inner Product Space and by Cauchy-Schwarz Inequality in a Complex Inner Product Space,

so the constant sequence $\langle\zeta,\widehat{u_{k}}\rangle=0$ converges in $\mathbb{C}$ (with the metric $|z-w|$) both to $0$ and to $\langle\zeta,\zeta\rangle$. By Uniqueness of Limits in a Metric Space, $\langle\zeta,\zeta\rangle=0$, and $\zeta=0$ by claim 4 of Elementary Properties of a Complex Inner Product.

1. (Algebra). Recall from The Tracial Algebra of a Noncommutative Law and Its Trace §algebra that $\mathcal{M}$ consists of the $T\in\mathcal{L}(\mathcal{H})$ with $TR_{p}=R_{p}T$ for all $p\in\mathcal{P}_{d}$. Clearly $IR_{p}=R_{p}=R_{p}I$, and $L_{p}R_{q}=R_{q}L_{p}$ by Left and Right Multiplication Operators on the Complex GNS Space of a Noncommutative Law: Boundedness, Algebra Rules, Adjoints, Commutation, the Vacuum and the Conjugation §commute; so $I,L_{p}\in\mathcal{M}$. Let $S,T\in\mathcal{M}$, $c\in\mathbb{C}$ and $p\in\mathcal{P}_{d}$. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §operations, $S+T,cS,ST\in\mathcal{L}(\mathcal{H})$ and

For the adjoint: $R_{p}^{*}=R_{p^{*}}$ by Left and Right Multiplication Operators on the Complex GNS Space of a Noncommutative Law: Boundedness, Algebra Rules, Adjoints, Commutation, the Vacuum and the Conjugation §adjoint, and $R_{p}$ is the adjoint of $R_{p}^{*}=R_{p^{*}}$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus, so $(R_{p^{*}})^{*}=R_{p}$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-unique. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus the composite $R_{p^{*}}S$ has the adjoint $S^{*}R_{p}$ and $SR_{p^{*}}$ has the adjoint $R_{p}S^{*}$. Since $R_{p^{*}}S=SR_{p^{*}}$, uniqueness of adjoints (Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-unique) gives $S^{*}R_{p}=R_{p}S^{*}$; as $S^{*}\in\mathcal{L}(\mathcal{H})$, $S^{*}\in\mathcal{M}$.

Now let $T_{k}\in\mathcal{M}$ with $T_{k}\to T$ in operator norm, $T\in\mathcal{L}(\mathcal{H})$. Using $T_{k}R_{p}=R_{p}T_{k}$ and the operations clause, $TR_{p}-R_{p}T=(T-T_{k})R_{p}+R_{p}(T_{k}-T)$, hence by the norm inequalities of Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §operations

for every $k$, using $T-T_{k}=(-1)(T_{k}-T)$. The right side tends to $0$, so the nonnegative number $\lVert TR_{p}-R_{p}T\rVert_{\mathrm{op}}$ is at most every positive real and hence is $0$ by Comparison of Real Numbers with Arbitrary Positive Slack §vanishing. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound, $0$ is then a bound for $TR_{p}-R_{p}T$, so $\lVert(TR_{p}-R_{p}T)v\rVert=0$ and $(TR_{p}-R_{p}T)v=0$ for every $v\in\mathcal{H}$ (claim 4 of Elementary Properties of a Complex Inner Product). Thus $TR_{p}=R_{p}T$ for every $p$, and $T\in\mathcal{M}$.

2. (Vacuum vectors). Let $T\in\mathcal{M}$ and $q\in\mathcal{P}_{d}$. Since $R_{q}\Omega=\widehat{q}$ by Left and Right Multiplication Operators on the Complex GNS Space of a Noncommutative Law: Boundedness, Algebra Rules, Adjoints, Commutation, the Vacuum and the Conjugation §vacuum, $T\widehat{q}=TR_{q}\Omega=R_{q}T\Omega$.

By Left and Right Multiplication Operators on the Complex GNS Space of a Noncommutative Law: Boundedness, Algebra Rules, Adjoints, Commutation, the Vacuum and the Conjugation §conjugation, $\langle J\xi,J\eta\rangle=\langle\eta,\xi\rangle$ and $JJ\eta=\eta$ for all $\xi,\eta\in\mathcal{H}$; applying the first with $J\eta$ in place of $\eta$ gives

Using the conventions, $R_{q}^{*}=R_{q^{*}}$ (Left and Right Multiplication Operators on the Complex GNS Space of a Noncommutative Law: Boundedness, Algebra Rules, Adjoints, Commutation, the Vacuum and the Conjugation §adjoint), $R_{q^{*}}\Omega=\widehat{q^{*}}=J\widehat{q}$ (Left and Right Multiplication Operators on the Complex GNS Space of a Noncommutative Law: Boundedness, Algebra Rules, Adjoints, Commutation, the Vacuum and the Conjugation §vacuum, Left and Right Multiplication Operators on the Complex GNS Space of a Noncommutative Law: Boundedness, Algebra Rules, Adjoints, Commutation, the Vacuum and the Conjugation §conjugation) and $(\ast\ast)$,

By claims 1 and 2 of Elementary Properties of a Complex Inner Product, $\langle T^{*}\Omega-JT\Omega,\widehat{q}\rangle=0$ for every $q$, so $T^{*}\Omega=JT\Omega$ by (P2).

If moreover $T\Omega=0$, then $T\widehat{q}=R_{q}0=0$ for every $q$, so $T$ and the zero map of $\mathcal{L}(\mathcal{H})$ agree on all classes, and $T=0$ by (P1).

3. (Bounded vectors). Define $\phi:\mathcal{P}_{d}\to\mathcal{H}$ by $\phi(q)=R_{q}\zeta$. By Left and Right Multiplication Operators on the Complex GNS Space of a Noncommutative Law: Boundedness, Algebra Rules, Adjoints, Commutation, the Vacuum and the Conjugation §algebra, $R_{q+q'}=R_{q}+R_{q'}$ and $R_{cq}=cR_{q}$, so $\phi$ is complex-linear. By hypothesis and Left and Right Multiplication Operators on the Complex GNS Space of a Noncommutative Law: Boundedness, Algebra Rules, Adjoints, Commutation, the Vacuum and the Conjugation §vacuum,

By The Complex Hilbert Completion is a Complex Hilbert Space Containing a Dense Isometric Image, and Bounded Complex-Linear Maps Extend to It §extension-linear (with $V=\mathcal{P}_{d}$, $h=h_{\lambda}$, $K=\mathcal{H}$), there is $T\in\mathcal{L}(\mathcal{H})$ with bound $C$ and $T\widehat{q}=R_{q}\zeta$ for every $q$; hence $\lVert T\rVert_{\mathrm{op}}\le C$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound. For $p,q\in\mathcal{P}_{d}$, using $R_{p}\widehat{q}=\widehat{qp}$ (Left and Right Multiplication Operators on the Complex GNS Space of a Noncommutative Law: Boundedness, Algebra Rules, Adjoints, Commutation, the Vacuum and the Conjugation §multiplication) and $R_{qp}=R_{p}R_{q}$ (Left and Right Multiplication Operators on the Complex GNS Space of a Noncommutative Law: Boundedness, Algebra Rules, Adjoints, Commutation, the Vacuum and the Conjugation §algebra),

so $TR_{p}=R_{p}T$ by (P1), and $T\in\mathcal{M}$. Also $T\Omega=T\widehat{1}=R_{1}\zeta=\zeta$, using The Complex GNS Space of a Tracial State on Noncommutative Polynomials §vacuum and $R_{1}=I$ (Left and Right Multiplication Operators on the Complex GNS Space of a Noncommutative Law: Boundedness, Algebra Rules, Adjoints, Commutation, the Vacuum and the Conjugation §algebra). If $T'\in\mathcal{M}$ also satisfies $T'\Omega=\zeta$, then by claim 2, $T'\widehat{q}=R_{q}T'\Omega=R_{q}\zeta=T\widehat{q}$ for every $q$, so $T'=T$ by (P1).

4. (Trace). By The Tracial Algebra of a Noncommutative Law and Its Trace §trace, $\tau(T)=\langle\Omega,T\Omega\rangle$, which is linear in $T$ by conditions 2 and 3 of Complex Inner Product Space ($\mathcal{M}$ is closed under sums and scalar multiples by claim 1). By Left and Right Multiplication Operators on the Complex GNS Space of a Noncommutative Law: Boundedness, Algebra Rules, Adjoints, Commutation, the Vacuum and the Conjugation §vacuum, $\tau(I)=\langle\Omega,\Omega\rangle=\lVert\Omega\rVert^{2}=1$ and $\tau(L_{p})=\langle\Omega,L_{p}\Omega\rangle=\lambda(p)$.

Let $S,T\in\mathcal{M}$; then $ST,TS,T^{*},T^{*}T\in\mathcal{M}$ by claim 1. By the conventions, claim 2 and $(\ast\ast)$,

By the conventions and condition 1 of Complex Inner Product Space, $\tau(T^{*})=\langle\Omega,T^{*}\Omega\rangle=\langle T\Omega,\Omega\rangle=\overline{\langle\Omega,T\Omega\rangle}=\overline{\tau(T)}$. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus, $\tau(T^{*}T)=\langle\Omega,T^{*}T\Omega\rangle=\lVert T\Omega\rVert^{2}\ge0$. If $\tau(T^{*}T)=0$, then $\langle T\Omega,T\Omega\rangle=0$, so $T\Omega=0$ by claim 4 of Elementary Properties of a Complex Inner Product, and $T=0$ by claim 2.

5. (Square roots and positivity of the trace). Let $T\in\mathcal{M}$ with $T\ge0$. By Square Roots of Positive Bounded Operators on a Complex Hilbert Space, Commuting with Everything that Commutes with the Operator §square-root there is $S\in\mathcal{L}(\mathcal{H})$ with $S\ge0$ and $SS=T$, and by Square Roots of Positive Bounded Operators on a Complex Hilbert Space, Commuting with Everything that Commutes with the Operator §commutation $S$ commutes with every element of $\mathcal{L}(\mathcal{H})$ that commutes with $T$. Each $R_{p}$ commutes with $T$ because $T\in\mathcal{M}$, so $SR_{p}=R_{p}S$ for every $p$, and $S\in\mathcal{M}$.

Let $A,B\in\mathcal{M}$ with $A\ge0$ and $B\ge0$, and let $S\in\mathcal{M}$ with $S\ge0$ and $SS=A$ as just shown. Then $SB\in\mathcal{M}$ by claim 1, and by claim 4 applied to $S$ and $SB$,

the last step because $S$ is self-adjoint (Self-Adjoint Operator). Since $B$ is positive semi-definite (Positive Semi-Definite Operator), $\langle S\Omega,B(S\Omega)\rangle$ is a real number and is $\ge0$.

Finally let $T\in\mathcal{M}$ and $A=\lVert T\rVert_{\mathrm{op}}^{2}I-T^{*}T\in\mathcal{L}(\mathcal{H})$. By Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus, $I$ is its own adjoint and $T^{*}T$ is self-adjoint, hence its own adjoint by the same clause and Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-unique; so, again by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus, $A$ has the adjoint $\lVert T\rVert_{\mathrm{op}}^{2}I-T^{*}T=A$ (the scalar $\lVert T\rVert_{\mathrm{op}}^{2}$ being real), and $A$ is self-adjoint by the same clause. For $v\in\mathcal{H}$, by conditions 2 and 3 of Complex Inner Product Space and Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §adjoint-calculus,

a real number, which is $\ge0$ because $\lVert Tv\rVert\le\lVert T\rVert_{\mathrm{op}}\lVert v\rVert$ by Bounded Linear Maps between Complex Inner Product Spaces: the Least Bound, Operations, the Underlying Real Structure, Adjoints, Completeness and the Quadratic-Form Bound §least-bound. Thus $A$ is positive semi-definite, and $A\ge0$.